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Ananagrams |
Most crossword puzzle fans are used to anagrams--groups of words with the same letters in different orders--for example OPTS, SPOT, STOP, POTS and POST. Some words however do not have this attribute, no matter how you rearrange their letters, you cannot form another word. Such words are calledananagrams, an example is QUIZ.
Obviously such definitions depend on the domain within which we are working; you might think that ATHENE is an ananagram, whereas any chemist would quickly produce ETHANE. One possible domain would be the entire English language, but this could lead to some problems. One could restrict the domain to, say, Music, in which case SCALE becomes a relative ananagram (LACES is not in the same domain) but NOTE is not since it can produce TONE.
Write a program that will read in the dictionary of a restricted domain and determine the relative ananagrams. Note that single letter words are, ipso facto, relative ananagrams since they cannot be ``rearranged‘‘ at all. The dictionary will contain no more than 1000 words.
Input will consist of a series of lines. No line will be more than 80 characters long, but may contain any number of words. Words consist of up to 20 upper and/or lower case letters, and will not be broken across lines. Spaces may appear freely around words, and at least one space separates multiple words on the same line. Note that words that contain the same letters but of differing case are considered to be anagrams of each other, thus tIeD and EdiT are anagrams. The file will be terminated by a line consisting of a single #.
Output will consist of a series of lines. Each line will consist of a single word that is a relative ananagram in the input dictionary. Words must be output in lexicographic (case-sensitive) order. There will always be at least one relative ananagram.
ladder came tape soon leader acme RIDE lone Dreis peat ScAlE orb eye Rides dealer NotE derail LaCeS drIed noel dire Disk mace Rob dries #
Disk NotE derail drIed eye ladder soon
1 // UVa156 Ananagrams 2 // Rujia Liu 3 //题意:输入一些单词,找出所有满足如下条件的单词:该单词不能通过字母重排得到输入文本中的另外一个单词 4 // 算法:把每个单词“标准化”,即全部转化为小写字母然后排序,然后放到map中进行统计 5 //代码由刘汝佳提供,我是在我的水平下,加以注释,学习大神的思想 6 #include<iostream> 7 #include<string> 8 #include<cctype>//用tolower() 9 #include<vector> 10 #include<map> 11 #include<algorithm> 12 using namespace std; 13 14 map<string,int> cnt; 15 vector<string> words; 16 17 //将单词s进行“标准化” 18 string repr(string s) { 19 string ans = s; 20 for(int i = 0; i < ans.length(); i++) 21 ans[i] = tolower(ans[i]);//转化成小写 22 sort(ans.begin(), ans.end());//从小到大排序 23 return ans; 24 } 25 26 int main() { 27 int n = 0; 28 string s; 29 while(cin >> s) { 30 if(s[0] == ‘#‘) break; 31 words.push_back(s);//vector的添加元素 32 string r = repr(s); 33 if(!cnt.count(r)) cnt[r] = 0;//cnt.count(r)返回r出现次数 34 cnt[r]++; 35 } 36 vector<string> ans; 37 for(int i = 0; i < words.size(); i++) 38 if(cnt[repr(words[i])] == 1) ans.push_back(words[i]); 39 sort(ans.begin(), ans.end()); 40 for(int i = 0; i < ans.size(); i++) 41 cout << ans[i] << "\n"; 42 return 0; 43 }
注释应该没错,
但是map还有很多用法还不清楚,等有空专门写一下map的用法(针对C++零基础)
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原文地址:http://www.cnblogs.com/liangyongrui/p/4275995.html